Name 3 muscles of the leg

bearhunter [10]

Answer:

n physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion.[1] It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.

In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed v is {\displaystyle {\begin{smallmatrix}{\frac {1}{2}}mv^{2}\end{smallmatrix}}}{\begin{smallmatrix}{\frac {1}{2}}mv^{2}\end{smallmatrix}}. In relativistic mechanics, this is a good approximation only when v is much less than the speed of light.

The standard unit of kinetic energy is the joule, while the imperial unit of kinetic energy is the foot-pound.

Explanation:

6 0

3 years ago

Read 2 more answers

A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

8 0

3 years ago

Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (

Sauron [17]

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

2.9 kg (0, 3.2)

4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

$x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}$

$0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}$

$10 + 8.3 x = 0$

$x = -1.20 m$

Similarly for y direction we have

$y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}$

$0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}$

$9.28 + 8.3 x = 0$

$x = -1.12 m$

5 0

3 years ago

Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh

mel-nik [20]

Answer: $u=\frac{v}{2}\sqrt{5-4sin\phi }$

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

$m(2v)+m(-vsin\phi )=2m(ucos\theta )$

$2ucos\theta =v(2-sin\theta )$ ------1

Conserving momentum in x direction

$mvcos\phi =2musin\theta$ -----2

squaring and adding 1 &2

$(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2$

$4u^2=4v^2+v^2-4v^2sin\phi$

$4u^2=5v^2-4v^2sin\phi$

$u=\frac{v}{2}\sqrt{5-4sin\phi }$

7 0

2 years ago

Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20Â°C. Container A is rigid. Container B has a 100 cm^2 pis

Vikki [24]

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) / 0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

7 0

2 years ago

Name 3 muscles of the leg​

Name 3 muscles of the leg